Is normal-ordering useful in supersymmetric theories?
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What are the rules of normal ordering, and why is it used in quantum field theory?
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Prahar Mitra, Ph.D. Physics, Harvard University (2017)
Answered Nov 28, 2015 · Author has 166answers and 1m answer views
I assume from the wording of the question that the OP has some knowledge of QFT beyond that of a layman and hence answer as such.
A very important operation in quantum mechanics is taking product of field operators. For instance, given to operators ϕ1(x)ϕ1(x) and ϕ2(x)ϕ2(x), I would like to be able to define the product operator O(x)=ϕ1(x)ϕ2(x)O(x)=ϕ1(x)ϕ2(x). Typically, this product is impossible to define in QFT since it is infinite. In other words, correlation functions involving O(x)O(x) as defined above will typically be infinite. Therefore, such an operator is really ill-defined. To make these operator products well-defined, we define what is known as "normal ordering of operators". There is no unique way to normal order a product of two operators. However, all normal orderings should satisfy at least one property - the normal ordered product of operators should be finite. As long as this is satisfied, any choice of normal ordering is a valid choice.
For example, let ϕ(x)ϕ(x) be a canonically normalized free scalar field in four dimensions. Then, we can define a normal ordered product operator as
O(x)=limy→x[ϕ(x)ϕ(y)−14π(x−y)2]O(x)=limy→x[ϕ(x)ϕ(y)−14π(x−y)2]
With this definition, it will turn out that all correlation functions of O(x)O(x) is completely finite, thereby making it a well-defined operator.
A choice of normal ordering that you will often see in QFT, particularly in perturbation theory is the "creation-annihilation normal ordering". In this choice of normal ordering, on expands every field in terms of creation and annihilation operators (this can be done in perturbation theory only in the interaction picture). Then, the normal ordering is defined by "moving all annihilation operators to the right and all creation operators to the left". This also turns out to give operators that are well-defined (i.e. finite.).
In conformal field theories, another choice of normal ordering is often used - called conformal normal ordering. Here, we define the product operator by subtracting off all the singular terms in the OPE of the operators.
Finally, to answer the OPs question about distributing it over addition. You can distribute it over addition as long as the components themselves are finite and defined.
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2 ANSWERS

Prahar Mitra, Ph.D. Physics, Harvard University (2017)
Answered Nov 28, 2015 · Author has 166answers and 1m answer views
I assume from the wording of the question that the OP has some knowledge of QFT beyond that of a layman and hence answer as such.
A very important operation in quantum mechanics is taking product of field operators. For instance, given to operators ϕ1(x)ϕ1(x) and ϕ2(x)ϕ2(x), I would like to be able to define the product operator O(x)=ϕ1(x)ϕ2(x)O(x)=ϕ1(x)ϕ2(x). Typically, this product is impossible to define in QFT since it is infinite. In other words, correlation functions involving O(x)O(x) as defined above will typically be infinite. Therefore, such an operator is really ill-defined. To make these operator products well-defined, we define what is known as "normal ordering of operators". There is no unique way to normal order a product of two operators. However, all normal orderings should satisfy at least one property - the normal ordered product of operators should be finite. As long as this is satisfied, any choice of normal ordering is a valid choice.
For example, let ϕ(x)ϕ(x) be a canonically normalized free scalar field in four dimensions. Then, we can define a normal ordered product operator as
O(x)=limy→x[ϕ(x)ϕ(y)−14π(x−y)2]O(x)=limy→x[ϕ(x)ϕ(y)−14π(x−y)2]
With this definition, it will turn out that all correlation functions of O(x)O(x) is completely finite, thereby making it a well-defined operator.
A choice of normal ordering that you will often see in QFT, particularly in perturbation theory is the "creation-annihilation normal ordering". In this choice of normal ordering, on expands every field in terms of creation and annihilation operators (this can be done in perturbation theory only in the interaction picture). Then, the normal ordering is defined by "moving all annihilation operators to the right and all creation operators to the left". This also turns out to give operators that are well-defined (i.e. finite.).
In conformal field theories, another choice of normal ordering is often used - called conformal normal ordering. Here, we define the product operator by subtracting off all the singular terms in the OPE of the operators.
Finally, to answer the OPs question about distributing it over addition. You can distribute it over addition as long as the components themselves are finite and defined.
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6D, N = (1,0) supersymmetric theories look very similar to their 4D, N = 2 counterparts, there is an essential difference between the two: in the generic case 6D, N = (1,0).
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